FALSE: every finite group is a direct product of cyclic prime-power groups

Given: The objects and hypotheses in the statement.

Let $X$ be a set. A permutation of $X$ is a bijection $f : X \to X$ (def-injection-surjection-bijection). The symmetric group of $X$ is the set of all permutations of $X$, $\operatorname{Sym}(X) \;:=\; \{\, f : X \to X \;:\; f \text{ is a bijection} \,\},$ equipped with composition as its operation, $(f \circ g)(x) \;=\; f(g(x)) \qquad (x \in X),$ and with the identity map $\mathrm{id}_X$, given by $\mathrm{id}_X(x) = x$, as distinguished element. Composition of two bijections of $X$ is again a bijection of $X$ (def-injection-surjection-bijection), so $\operatorname{Sym}(X)$ is closed under $\circ$ and $\circ$ is a binary operation on it (def-binary-operation); and $\mathrm{id}_X$ is a bijection of $X$, so it is an element of $\operatorname{Sym}(X)$, and it is a two-sided identity for composition (def-identity-element) because $f \circ \mathrm{id}_X = f = \mathrm{id}_X \circ f$ holds pointwise for every $f$. That $(\operatorname{Sym}(X), \circ, \mathrm{id}_X)$ is a group is lem-symmetric-group-is-a-group. Cycle notation for a finite list of distinct points. For distinct elements $x_0, x_1, \dots, x_{k-1}$ of $X$ with $k \ge 2$, the symbol $(x_0\,x_1\,\cdots\,x_{k-1})$ denotes the permutation sending $x_i$ to $x_{i+1}$ for $i < k-1$, sending $x_{k-1}$ to $x_0$, and fixing every element of $X$ outside $\{x_0,\dots,x_{k-1}\}$. It is a bijection, because the map described sends the set $\{x_0,\dots,x_{k-1}\}$ onto itself by a rule with an evident inverse (send each $x_{i+1}$ back to $x_i$ and $x_0$ back to $x_{k-1}$) and fixes the complement pointwise. A transposition is such a symbol with $k = 2$, that is $(a\,b)$ with $a \ne b$: it exchanges $a$ and $b$ and fixes everything else, and it satisfies $(a\,b) \circ (a\,b) = \mathrm{id}_X$. A product of cycle symbols means their composite, so $(a\,b)(c\,d)$ is $(a\,b) \circ (c\,d)$. (The symmetric group $\operatorname{Sym}(X)$: the bijections of a set $X$ under composition).

For every set $X$, the triple $(\operatorname{Sym}(X), \circ, \mathrm{id}_X)$ of def-symmetric-group is a group (def-group); the inverse of a permutation $f$ is its inverse function $f^{-1}$. If $X$ contains three distinct elements $a$, $b$, $c$, then $\operatorname{Sym}(X)$ is not abelian: the transpositions $\tau = (a\,b)$ and $\rho = (b\,c)$ satisfy $\tau \circ \rho \ne \rho \circ \tau$. ($\operatorname{Sym}(X)$ is a group under composition, and it is non-abelian whenever $X$ has at least three distinct elements).

Let $A$ be a finite set with $n := \lvert A\rvert$ and write $\operatorname{Bij}(A) := \{\, f : A \to A \ :\ f \text{ is a bijection} \,\}.$ Then $\operatorname{Bij}(A)$ is finite and $\lvert\operatorname{Bij}(A)\rvert = n!$ (def-factorial-and-falling-factorial). More generally, for finite sets $X$ and $Y$ write $\operatorname{Bij}(X,Y)$ for the set of bijections $X \to Y$. If $\lvert X\rvert = \lvert Y\rvert = n$ then $\operatorname{Bij}(X,Y)$ is finite with $n!$ elements, and if $\lvert X\rvert \ne \lvert Y\rvert$ then $\operatorname{Bij}(X,Y) = \varnothing$. (A finite set $A$ with $\lvert A\rvert = n$ has exactly $n!$ bijections onto itself, and $n!$ bijections onto any set of the same cardinality).

Let $G$ be a group and $g \in G$, with integer powers as in def-group-power. Then $\langle g \rangle \;=\; \{\, g^{n} \;:\; n \in \mathbb{Z} \,\} ,$ the cyclic subgroup generated by $g$ (def-generated-subgroup) being exactly the set of integer powers of $g$. Consequently every cyclic group is abelian, and so is every cyclic subgroup of any group. ($\langle g \rangle = \{\, g^{n} : n \in \mathbb{Z} \,\}$, and every cyclic group is abelian).

Let $G$ and $H$ be groups. Their external direct product has underlying set $G\times H:=\{(g,h):g\in G,\ h\in H\}$ and componentwise operation $(g,h)(g',h') := (gg',hh').$ The fact that this operation makes $G\times H$ a group, with the indicated identity and inverses, is proved in thm-external-direct-product-is-a-group. Until that result is used, this definition introduces only the set and its componentwise binary operation. (The external direct product $G\times H$ with componentwise multiplication).

For groups $G$ and $H$, the componentwise operation of def-external-direct-product-of-groups makes $G\times H$ a group. Its identity is $(e_G,e_H)$, and $(g,h)^{-1}=(g^{-1},h^{-1}).$ Moreover the coordinate maps $\pi_G(g,h)=g$ and $\pi_H(g,h)=h$ are group homomorphisms. ($G\times H$ is a group with identity $(e_G,e_H)$, coordinatewise inverses, and homomorphic coordinate projections).